Minimal Polynomial
Cayley-Hamilton Theorem
Every square matrix satisfies its own characteristic equation.
A n + d 1 A n − 1 + ⋯ + d n − 1 A + d n I = 0 \mathbf{A}^n + d_1\mathbf{A}^{n-1} + \cdots + d_{n-1}\mathbf{A} + d_n\mathbf{I} = 0 A n + d 1 ​ A n − 1 + ⋯ + d n − 1 ​ A + d n ​ I = 0
Remark: Cayley-Hamilton Theorem basically states that nth and higher powers of a matrix can be expressed in terms of lower powers of the matrix. This is a very useful property in solving linear systems.
A n = − d 1 A n − 1 − ⋯ − d n − 1 A − d n I A^n = -d_1A^{n-1} - \cdots - d_{n-1}A - d_nI A n = − d 1 ​ A n − 1 − ⋯ − d n − 1 ​ A − d n ​ I
For previous example, we have:
A 2 − 4 A + 3 I = 0 A^2 - 4A + 3I = 0 A 2 − 4 A + 3 I = 0
A 2 = 4 A − 3 I A^2 = 4A - 3I A 2 = 4 A − 3 I
A 3 = 4 A 2 − 3 A = 4 ( 4 A − 3 I ) − 3 A = 13 A − 12 I A^3 = 4A^2 - 3A = 4(4A - 3I) - 3A = 13A - 12I A 3 = 4 A 2 − 3 A = 4 ( 4 A − 3 I ) − 3 A = 13 A − 12 I
A 4 = 4 A 3 − 3 A 2 = 4 ( 13 A − 12 I ) − 3 ( 4 A − 3 I ) = 49 A − 48 I A^4 = 4A^3 - 3A^2 = 4(13A - 12I) - 3(4A - 3I) = 49A - 48I A 4 = 4 A 3 − 3 A 2 = 4 ( 13 A − 12 I ) − 3 ( 4 A − 3 I ) = 49 A − 48 I
e x p ( s ) = ∑ n = 0 ∞ s n n ! = I + s + s 2 2 ! + s 3 3 ! + ⋯ \large exp(s) = \sum_{n=0}^{\infty} \frac{s^n}{n!} = I + s + \frac{s^2}{2!} + \frac{s^3}{3!} + \cdots e x p ( s ) = ∑ n = 0 ∞ ​ n ! s n ​ = I + s + 2 ! s 2 ​ + 3 ! s 3 ​ + ⋯
e x p ( A ) = ∑ n = 0 ∞ A n n ! = I + A + A 2 2 ! + A 3 3 ! + ⋯ \large exp(A) = \sum_{n=0}^{\infty} \frac{A^n}{n!} = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots e x p ( A ) = ∑ n = 0 ∞ ​ n ! A n ​ = I + A + 2 ! A 2 ​ + 3 ! A 3 ​ + ⋯
A − 1 = ? A^{-1} = ? A − 1 = ?
A 2 − 4 A + 3 I = 0 A^2 - 4A + 3I = 0 A 2 − 4 A + 3 I = 0
A − 4 I + 3 A − 1 = 0 A - 4I + 3A^{-1} = 0 A − 4 I + 3 A − 1 = 0
A − 1 = ( 4 I − A ) 3 A^{-1} = \Large \frac{(4I - A)}{3} A − 1 = 3 ( 4 I − A ) ​
Example: A = [ 2 1 1 2 ] A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} A = [ 2 1 ​ 1 2 ​ ] A 100 = ? A^{100}=? A 100 = ?
Solution:
A 100 = α A + β I A^{100} = \alpha A + \beta I A 100 = α A + β I
λ 1 = 3 , λ 2 = 1 , and e 1 = [ 1 1 ] , e 2 = [ 1 − 1 ] \lambda_1 = 3, \lambda_2 = 1 \text{, and } e_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, e_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} λ 1 ​ = 3 , λ 2 ​ = 1 , and e 1 ​ = [ 1 1 ​ ] , e 2 ​ = [ 1 − 1 ​ ]
A e 1 = λ 1 e 1 ⇒ A 2 e 1 = λ 1 2 e 1 ⇒ A 3 e 1 = λ 1 3 e 1 ⇒ ⋯ ⇒ A 100 e 1 = λ 1 100 e 1 A e_1 = \lambda_1 e_1 \Rightarrow A^2 e_1 = \lambda_1^2 e_1 \Rightarrow A^3 e_1 = \lambda_1^3 e_1 \Rightarrow \cdots \Rightarrow A^{100} e_1 = \lambda_1^{100} e_1 A e 1 ​ = λ 1 ​ e 1 ​ ⇒ A 2 e 1 ​ = λ 1 2 ​ e 1 ​ ⇒ A 3 e 1 ​ = λ 1 3 ​ e 1 ​ ⇒ ⋯ ⇒ A 100 e 1 ​ = λ 1 100 ​ e 1 ​
A e 2 = λ 2 e 2 ⇒ A 2 e 2 = λ 2 2 e 2 ⇒ A 3 e 2 = λ 2 3 e 2 ⇒ ⋯ ⇒ A 100 e 2 = λ 2 100 e 2 A e_2 = \lambda_2 e_2 \Rightarrow A^2 e_2 = \lambda_2^2 e_2 \Rightarrow A^3 e_2 = \lambda_2^3 e_2 \Rightarrow \cdots \Rightarrow A^{100} e_2 = \lambda_2^{100} e_2 A e 2 ​ = λ 2 ​ e 2 ​ ⇒ A 2 e 2 ​ = λ 2 2 ​ e 2 ​ ⇒ A 3 e 2 ​ = λ 2 3 ​ e 2 ​ ⇒ ⋯ ⇒ A 100 e 2 ​ = λ 2 100 ​ e 2 ​
A 100 = α A + β I ⇒ A 100 e 1 = α A e 1 + β e 1 ⇒ λ 1 100 e 1 = α λ 1 e 1 + β e 1 ⇒ λ 1 100 = α λ 1 + β    ⟹    3 100 = 3 α + β A^{100} = \alpha A + \beta I \Rightarrow A^{100} e_1 = \alpha A e_1 + \beta e_1 \Rightarrow \lambda_1^{100} e_1 = \alpha \lambda_1 e_1 + \beta e_1 \Rightarrow \lambda_1^{100} = \alpha \lambda_1 + \beta \implies 3^{100} = 3\alpha + \beta A 100 = α A + β I ⇒ A 100 e 1 ​ = α A e 1 ​ + β e 1 ​ ⇒ λ 1 100 ​ e 1 ​ = α λ 1 ​ e 1 ​ + β e 1 ​ ⇒ λ 1 100 ​ = α λ 1 ​ + β ⟹ 3 100 = 3 α + β
A 100 = α A + β I ⇒ A 100 e 2 = α A e 2 + β e 2 ⇒ λ 2 100 e 2 = α λ 2 e 2 + β e 2 ⇒ λ 2 100 = α λ 2 + β    ⟹    1 100 = α + β A^{100} = \alpha A + \beta I \Rightarrow A^{100} e_2 = \alpha A e_2 + \beta e_2 \Rightarrow \lambda_2^{100} e_2 = \alpha \lambda_2 e_2 + \beta e_2 \Rightarrow \lambda_2^{100} = \alpha \lambda_2 + \beta \implies 1^{100} = \alpha + \beta A 100 = α A + β I ⇒ A 100 e 2 ​ = α A e 2 ​ + β e 2 ​ ⇒ λ 2 100 ​ e 2 ​ = α λ 2 ​ e 2 ​ + β e 2 ​ ⇒ λ 2 100 ​ = α λ 2 ​ + β ⟹ 1 100 = α + β
{ 3 100 = 3 α + β 1 = α + β ⇒ α = 3 100 − 1 2 β = − 3 100 + 3 2 \begin{cases} 3^{100} = 3\alpha + \beta \\ 1 = \alpha + \beta \end{cases} \Rightarrow \begin{matrix} \alpha = \frac{3^{100} - 1}{2} \\ \beta = \frac{-3^{100} + 3}{2} \end{matrix} { 3 100 = 3 α + β 1 = α + β ​ ⇒ α = 2 3 100 − 1 ​ β = 2 − 3 100 + 3 ​ ​
A 100 = 3 100 − 1 2 A + − 3 100 + 3 2 I A^{100} = \frac{3^{100} - 1}{2} A + \frac{-3^{100} + 3}{2} I A 100 = 2 3 100 − 1 ​ A + 2 − 3 100 + 3 ​ I
Minimal Polynomial
Definition: The monic polynomial is the polynomial with the highest degree coefficient equal to 1.
s n + a 1 s n − 1 + ⋯ → s^n + a_1s^{n-1} + \cdots \rightarrow s n + a 1 ​ s n − 1 + ⋯ → monic
2 s n + a 1 s n − 1 + ⋯ → 2s^n + a_1s^{n-1} + \cdots \rightarrow 2 s n + a 1 ​ s n − 1 + ⋯ → not monic
Definition: The minimal polynomial of a matrix is the monic polynomial of the lowest degree that has the matrix as a root.
m ( A ) = 0 n × n m(A) = 0_{n \times n} m ( A ) = 0 n × n ​
Theorem: Given a matrix A = C n  t i m e s n A = \mathbb{C}^{n \ times n} A = C n  t im es n , let m(s) be its minimal polynomial. Then, the minimal polynomial is the smallest degree polynomial that satisfies the following:
i) m(s) is unique.
ii) m(s) divides d(s) with no remainder.
∃ q ( s )  s.t. d ( s ) = q ( s ) m ( s ) \exists q(s) \text{ s.t. } d(s) = q(s)m(s) ∃ q ( s )  s.t. d ( s ) = q ( s ) m ( s )
iii) Every root of m(s) is a root of d(s).
Example: A = [ 1 0 0 0 2 0 0 0 3 ] A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} A = ⎣ ⎡ ​ 1 0 0 ​ 0 2 0 ​ 0 0 3 ​ ⎦ ⎤ ​ d ( s ) = ( s − 1 ) ( s − 2 ) ( s − 3 ) m ( s ) = ( s − 1 ) ( s − 2 ) ( s − 3 ) \Large \substack{d(s) = (s-1)(s-2)(s-3) \\ m(s) = (s-1)(s-2)(s-3)} d ( s ) = ( s − 1 ) ( s − 2 ) ( s − 3 ) m ( s ) = ( s − 1 ) ( s − 2 ) ( s − 3 ) ​
Remark: When A has distinct eigenvalues (which implies diagonalizability but converse is not true), the minimal polynomial is the same as the characteristic polynomial.
Example: A 1 = [ 1 0 0 0 2 0 0 0 2 ] A_1 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} A 1 ​ = ⎣ ⎡ ​ 1 0 0 ​ 0 2 0 ​ 0 0 2 ​ ⎦ ⎤ ​ d ( s ) = ( s − 1 ) ( s − 2 ) 2 m ( s ) = ( s − 1 ) ( s − 2 )        A 2 = [ 1 0 0 0 2 1 0 0 2 ] \Large \substack{d(s) = (s-1)(s-2)^2 \\ m(s) = (s-1)(s-2)} \ \ \ \ \ \ \ \normalsize A_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} d ( s ) = ( s − 1 ) ( s − 2 ) 2 m ( s ) = ( s − 1 ) ( s − 2 ) ​        A 2 ​ = ⎣ ⎡ ​ 1 0 0 ​ 0 2 0 ​ 0 1 2 ​ ⎦ ⎤ ​ d ( s ) = ( s − 1 ) ( s − 2 ) 2 m ( s ) = ( s − 1 ) ( s − 2 ) 2 \Large \substack{d(s) = (s-1)(s-2)^2 \\ m(s) = (s-1)(s-2)^2} d ( s ) = ( s − 1 ) ( s − 2 ) 2 m ( s ) = ( s − 1 ) ( s − 2 ) 2 ​
A way to check if a minimal polynomial is correct is to check if the characteristic polynomial is zero when the minimal polynomial is substituted for s.
Let m 1 ( s ) = ( s − 1 ) ( s − 2 ) m_1(s) = (s-1)(s-2) m 1 ​ ( s ) = ( s − 1 ) ( s − 2 ) then m 1 ( A ) = ( A − 1 I ) ( A − 2 I ) = [ 0 0 0 0 1 0 0 0 1 ] [ − 1 0 0 0 0 0 0 0 0 ] = [ 0 0 0 0 0 0 0 0 0 ] ✓ m_1(A) = (A-1I)(A-2I) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \checkmark m 1 ​ ( A ) = ( A − 1 I ) ( A − 2 I ) = ⎣ ⎡ ​ 0 0 0 ​ 0 1 0 ​ 0 0 1 ​ ⎦ ⎤ ​ ⎣ ⎡ ​ − 1 0 0 ​ 0 0 0 ​ 0 0 0 ​ ⎦ ⎤ ​ = ⎣ ⎡ ​ 0 0 0 ​ 0 0 0 ​ 0 0 0 ​ ⎦ ⎤ ​ ✓
Let m 2 ( s ) = ( s − 1 ) ( s − 2 ) m_2(s) = (s-1)(s-2) m 2 ​ ( s ) = ( s − 1 ) ( s − 2 ) then m 2 ( A ) = ( A − 1 I ) ( A − 2 I ) = [ 0 0 0 0 1 0 0 0 1 ] [ − 1 0 0 0 0 1 0 0 0 ] = [ 0 0 0 0 0 1 0 0 0 ] ≠0 m_2(A) = (A-1I)(A-2I) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \neq 0 m 2 ​ ( A ) = ( A − 1 I ) ( A − 2 I ) = ⎣ ⎡ ​ 0 0 0 ​ 0 1 0 ​ 0 0 1 ​ ⎦ ⎤ ​ ⎣ ⎡ ​ − 1 0 0 ​ 0 0 0 ​ 0 1 0 ​ ⎦ ⎤ ​ = ⎣ ⎡ ​ 0 0 0 ​ 0 0 0 ​ 0 1 0 ​ ⎦ ⎤ ​ î€ = 0
Then, m 2 ( s ) = ( s − 1 ) ( s − 2 ) 2    ⟹    m 2 ( A ) = ( A − 1 I ) ( A − 2 I ) 2 = [ 0 0 0 0 1 0 0 0 1 ] [ − 1 0 0 0 0 1 0 0 0 ] [ − 1 0 0 0 0 1 0 0 0 ] = [ 0 0 0 0 0 0 0 0 0 ] ✓ m_2(s) = (s-1)(s-2)^2 \implies m_2(A) = (A-1I)(A-2I)^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \checkmark m 2 ​ ( s ) = ( s − 1 ) ( s − 2 ) 2 ⟹ m 2 ​ ( A ) = ( A − 1 I ) ( A − 2 I ) 2 = ⎣ ⎡ ​ 0 0 0 ​ 0 1 0 ​ 0 0 1 ​ ⎦ ⎤ ​ ⎣ ⎡ ​ − 1 0 0 ​ 0 0 0 ​ 0 1 0 ​ ⎦ ⎤ ​ ⎣ ⎡ ​ − 1 0 0 ​ 0 0 0 ​ 0 1 0 ​ ⎦ ⎤ ​ = ⎣ ⎡ ​ 0 0 0 ​ 0 0 0 ​ 0 0 0 ​ ⎦ ⎤ ​ ✓
R 3 = N ( A − 1 I ) ⊕ N ( ( A − 2 I ) 2 ) \mathbb{R}^3 = N(A-1I) \oplus N((A-2I)^2) R 3 = N ( A − 1 I ) ⊕ N (( A − 2 I ) 2 ) where,
N ( A − 1 I ) = N ( [ 0 0 0 0 1 0 0 0 1 ] ) = span { [ 1 0 0 ] }   dim ( N ( A − 1 I ) ) = 1 N(A-1I) = N \bigg ( \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \bigg ) = \text{span} \bigg \{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \bigg \} \ \ \text{dim} \bigg ( N(A-1I) \bigg ) = 1 N ( A − 1 I ) = N ( ⎣ ⎡ ​ 0 0 0 ​ 0 1 0 ​ 0 0 1 ​ ⎦ ⎤ ​ ) = span { ⎣ ⎡ ​ 1 0 0 ​ ⎦ ⎤ ​ }   dim ( N ( A − 1 I ) ) = 1
N ( ( A − 2 I ) 2 ) = N ( [ 1 0 0 0 0 0 0 0 0 ] ) = span { [ 0 1 0 ] , [ 0 0 1 ] }   dim ( N ( ( A − 2 I ) 2 ) ) = 2 N((A-2I)^2) = N \bigg ( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \bigg ) = \text{span} \bigg \{ \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg \} \ \ \text{dim} \bigg ( N((A-2I)^2) \bigg ) = 2 N (( A − 2 I ) 2 ) = N ( ⎣ ⎡ ​ 1 0 0 ​ 0 0 0 ​ 0 0 0 ​ ⎦ ⎤ ​ ) = span { ⎣ ⎡ ​ 0 1 0 ​ ⎦ ⎤ ​ , ⎣ ⎡ ​ 0 0 1 ​ ⎦ ⎤ ​ }   dim ( N (( A − 2 I ) 2 ) ) = 2
Remark: There is a relationship between decomposition form and minimal polynomial in terms of power.
m ( s ) = ( s − λ 1 ) d 1 ⋯ ( s − λ k ) d k m(s) = (s-\lambda_1)^{d_1} \cdots (s-\lambda_k)^{d_k} m ( s ) = ( s − λ 1 ​ ) d 1 ​ ⋯ ( s − λ k ​ ) d k ​
R n = N ( A − λ 1 I ) d 1 ⊕ ⋯ ⊕ N ( A − λ k I ) d k \mathbb{R}^n = N(A-\lambda_1I)^{d_1} \oplus \cdots \oplus N(A-\lambda_kI)^{d_k} R n = N ( A − λ 1 ​ I ) d 1 ​ ⊕ ⋯ ⊕ N ( A − λ k ​ I ) d k ​
Theorem:
C n = N ( A − λ 1 I ) m 1 ⊕ N ( A − λ 2 I ) m 2 ⊕ ⋯ ⊕ N ( A − λ k I ) m k \mathbb{C}^n = N(A-\lambda_1I)^{m_1} \oplus N(A-\lambda_2I)^{m_2} \oplus \cdots \oplus N(A-\lambda_kI)^{m_k} C n = N ( A − λ 1 ​ I ) m 1 ​ ⊕ N ( A − λ 2 ​ I ) m 2 ​ ⊕ ⋯ ⊕ N ( A − λ k ​ I ) m k ​
d ( s ) = ( s − λ 1 ) r 1 ⋯ ( s − λ k ) r k d(s) = (s-\lambda_1)^{r_1} \cdots (s-\lambda_k)^{r_k} d ( s ) = ( s − λ 1 ​ ) r 1 ​ ⋯ ( s − λ k ​ ) r k ​
r 1 + r 2 + ⋯ + r k = n r_1 + r_2 + \cdots + r_k = n r 1 ​ + r 2 ​ + ⋯ + r k ​ = n
m ( s ) = ( s − λ 1 ) m 1 ⋯ ( s − λ k ) m k m(s) = (s-\lambda_1)^{m_1} \cdots (s-\lambda_k)^{m_k} m ( s ) = ( s − λ 1 ​ ) m 1 ​ ⋯ ( s − λ k ​ ) m k ​
1 ≤ m i ≤ r i 1 \leq m_i \leq r_i 1 ≤ m i ​ ≤ r i ​
A ˉ = [ A ˉ 1 0 ⋯ 0 0 A ˉ 2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ A ˉ k ] \bar A = \begin{bmatrix} \bar A_1 & 0 & \cdots & 0 \\ 0 & \bar A_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \bar A_k \end{bmatrix} A ˉ = ⎣ ⎡ ​ A ˉ 1 ​ 0 ⋮ 0 ​ 0 A ˉ 2 ​ ⋮ 0 ​ ⋯ ⋯ ⋱ ⋯ ​ 0 0 ⋮ A ˉ k ​ ​ ⎦ ⎤ ​ A ˉ = B − 1 A B , B is composed of the basis vectors B = [ e 1 e 2 ⋯ e n ]  for the N ( A − λ i I ) m i \Large \substack{ \bar A = B^{-1}AB \text{, B is composed of the basis vectors} \\ B = \begin{bmatrix} e_1 & e_2 & \cdots & e_n \end{bmatrix} \text{ for the } N(A-\lambda_iI)^{m_i} } A ˉ = B − 1 A B , B is composed of the basis vectors B = [ e 1 ​ ​ e 2 ​ ​ ⋯ ​ e n ​ ​ ]  for the N ( A − λ i ​ I ) m i ​ ​
Size of A ˉ i \bar A_i A ˉ i ​ is dim ( N ( A − λ i I ) m i ) \text{dim} \bigg ( N(A-\lambda_iI)^{m_i} \bigg ) dim ( N ( A − λ i ​ I ) m i ​ )
Example: A = [ 1 0 0 0 0 1 1 0 0 0 1 0 0 0 0 2 ] A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix} A = ⎣ ⎡ ​ 1 0 0 0 ​ 0 1 0 0 ​ 0 1 1 0 ​ 0 0 0 2 ​ ⎦ ⎤ ​ d ( s ) = ( s − 1 ) 3 ( s − 2 ) m ( s ) = ( s − 1 ) 2 ( s − 2 ) \Large \substack{d(s) = (s-1)^3(s-2) \\ m(s) = (s-1)^2(s-2)} d ( s ) = ( s − 1 ) 3 ( s − 2 ) m ( s ) = ( s − 1 ) 2 ( s − 2 ) ​
Solution: Let Σ 1 \Sigma_1 Σ 1 ​ be A − λ 1 I A-\lambda_1I A − λ 1 ​ I and Σ 2 \Sigma_2 Σ 2 ​ be A − λ 2 I A-\lambda_2I A − λ 2 ​ I .
Σ 1 = [ 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 ] dim ( N ( Σ 1 ) ) = 2 ≠3 = r 1 → dim V − dim R ( Σ 1 ) = 4 − 2 = 2 \Sigma_1 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1) \bigg ) = 2 \neq 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1)}= 4 - 2 = 2 Σ 1 ​ = ⎣ ⎡ ​ 0 0 0 0 ​ 0 0 0 0 ​ 0 1 0 0 ​ 0 0 0 1 ​ ⎦ ⎤ ​ dim ( N ( Σ 1 ​ ) ) = 2 î€ = 3 = r 1 ​ → dim V − dim R ( Σ 1 ​ ) = 4 − 2 = 2
Then we need to check the dimension of N ( Σ 1 2 ) N(\Sigma_1^2) N ( Σ 1 2 ​ ) .
Σ 1 2 = [ 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 ] [ 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 ] = [ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 ] dim ( N ( Σ 1 2 ) ) = 1 = r 1 → dim V − dim R ( Σ 1 2 ) = 4 − 1 = 3 = r 1 \Sigma_1^2 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^2) \bigg ) = 1 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^2)} = 4 - 1 = 3 = r_1 Σ 1 2 ​ = ⎣ ⎡ ​ 0 0 0 0 ​ 0 0 0 0 ​ 0 1 0 0 ​ 0 0 0 1 ​ ⎦ ⎤ ​ ⎣ ⎡ ​ 0 0 0 0 ​ 0 0 0 0 ​ 0 1 0 0 ​ 0 0 0 1 ​ ⎦ ⎤ ​ = ⎣ ⎡ ​ 0 0 0 0 ​ 0 0 0 0 ​ 0 0 0 0 ​ 0 0 0 1 ​ ⎦ ⎤ ​ dim ( N ( Σ 1 2 ​ ) ) = 1 = r 1 ​ → dim V − dim R ( Σ 1 2 ​ ) = 4 − 1 = 3 = r 1 ​
Lets briefly check the dimension of N ( Σ 1 3 ) N(\Sigma_1^3) N ( Σ 1 3 ​ ) .
Σ 1 3 = [ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 ] dim ( N ( Σ 1 3 ) ) = 3 = r 1 → dim V − dim R ( Σ 1 3 ) = 4 − 1 = 3 = r 1 \Sigma_1^3 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^3) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^3)} = 4 - 1 = 3 = r_1 Σ 1 3 ​ = ⎣ ⎡ ​ 0 0 0 0 ​ 0 0 0 0 ​ 0 0 0 0 ​ 0 0 0 1 ​ ⎦ ⎤ ​ dim ( N ( Σ 1 3 ​ ) ) = 3 = r 1 ​ → dim V − dim R ( Σ 1 3 ​ ) = 4 − 1 = 3 = r 1 ​
                      ⋮ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots                       ⋮
Σ 1 4 = [ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 ] dim ( N ( Σ 1 4 ) ) = 3 = r 1 → dim V − dim R ( Σ 1 4 ) = 4 − 1 = 3 = r 1 \Sigma_1^4 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^4) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^4)} = 4 - 1 = 3 = r_1 Σ 1 4 ​ = ⎣ ⎡ ​ 0 0 0 0 ​ 0 0 0 0 ​ 0 0 0 0 ​ 0 0 0 1 ​ ⎦ ⎤ ​ dim ( N ( Σ 1 4 ​ ) ) = 3 = r 1 ​ → dim V − dim R ( Σ 1 4 ​ ) = 4 − 1 = 3 = r 1 ​
Lets check the second eigenvalue.
Σ 2 = [ − 1 0 0 0 0 − 1 1 0 0 0 − 1 0 0 0 0 0 ] = [ 1 0 0 0 0 1 − 1 0 0 0 1 0 0 0 0 0 ] dim ( N ( Σ 2 ) ) = 1 = r 2 → dim V − dim R ( Σ 2 ) = 4 − 3 = 1 \Sigma_2 = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_2) \bigg ) = 1 = r_2 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_2)} = 4 - 3 = 1 Σ 2 ​ = ⎣ ⎡ ​ − 1 0 0 0 ​ 0 − 1 0 0 ​ 0 1 − 1 0 ​ 0 0 0 0 ​ ⎦ ⎤ ​ = ⎣ ⎡ ​ 1 0 0 0 ​ 0 1 0 0 ​ 0 − 1 1 0 ​ 0 0 0 0 ​ ⎦ ⎤ ​ dim ( N ( Σ 2 ​ ) ) = 1 = r 2 ​ → dim V − dim R ( Σ 2 ​ ) = 4 − 3 = 1
Then the minimal polynomial is m ( s ) = ( s − 1 ) 2 ( s − 2 ) m(s) = (s-1)^2(s-2) m ( s ) = ( s − 1 ) 2 ( s − 2 ) .
Example: A = [ 1 1 0 0 0 1 1 0 0 0 1 0 0 0 0 2 ] A = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix} A = ⎣ ⎡ ​ 1 0 0 0 ​ 1 1 0 0 ​ 0 1 1 0 ​ 0 0 0 2 ​ ⎦ ⎤ ​ d ( s ) = ( s − 1 ) 3 ( s − 2 ) m ( s ) = ( s − 1 ) 3 ( s − 2 ) \Large \substack{d(s) = (s-1)^3(s-2) \\ m(s) = (s-1)^3(s-2)} d ( s ) = ( s − 1 ) 3 ( s − 2 ) m ( s ) = ( s − 1 ) 3 ( s − 2 ) ​
Solution: Let Σ 1 \Sigma_1 Σ 1 ​ be A − λ 1 I A-\lambda_1I A − λ 1 ​ I and Σ 2 \Sigma_2 Σ 2 ​ be A − λ 2 I A-\lambda_2I A − λ 2 ​ I .
Σ 1 = [ 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 ] dim ( N ( Σ 1 ) ) = 1 ≠r 1 → dim V − dim R ( Σ 1 ) = 4 − 3 = 1 \Sigma_1 = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1) \bigg ) = 1 \neq r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1)} = 4 - 3 = 1 Σ 1 ​ = ⎣ ⎡ ​ 0 0 0 0 ​ 1 0 0 0 ​ 0 1 0 0 ​ 0 0 0 1 ​ ⎦ ⎤ ​ dim ( N ( Σ 1 ​ ) ) = 1 î€ = r 1 ​ → dim V − dim R ( Σ 1 ​ ) = 4 − 3 = 1
Then we need to check the dimension of N ( Σ 1 2 ) N(\Sigma_1^2) N ( Σ 1 2 ​ ) .
Σ 1 2 = [ 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 ] dim ( N ( Σ 1 2 ) ) = 2 ≠r 1 → dim V − dim R ( Σ 1 2 ) = 4 − 2 = 2 \Sigma_1^2 = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^2) \bigg ) = 2 \neq r_1 \rightarrow \text{dim}{V}- \text{dim}{R(\Sigma_1^2)} = 4 - 2 = 2 Σ 1 2 ​ = ⎣ ⎡ ​ 0 0 0 0 ​ 0 0 0 0 ​ 1 0 0 0 ​ 0 0 0 1 ​ ⎦ ⎤ ​ dim ( N ( Σ 1 2 ​ ) ) = 2 î€ = r 1 ​ → dim V − dim R ( Σ 1 2 ​ ) = 4 − 2 = 2
Check the dimension of N ( Σ 1 3 ) N(\Sigma_1^3) N ( Σ 1 3 ​ ) .
Σ 1 3 = [ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 ] dim ( N ( Σ 1 3 ) ) = 3 = r 1 → dim V − dim R ( Σ 1 3 ) = 4 − 1 = 3 \Sigma_1^3 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^3) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^3)}= 4 - 1 = 3 Σ 1 3 ​ = ⎣ ⎡ ​ 0 0 0 0 ​ 0 0 0 0 ​ 0 0 0 0 ​ 0 0 0 1 ​ ⎦ ⎤ ​ dim ( N ( Σ 1 3 ​ ) ) = 3 = r 1 ​ → dim V − dim R ( Σ 1 3 ​ ) = 4 − 1 = 3
Now we can check the second eigenvalue.
Σ 2 = [ − 1 1 0 0 0 − 1 1 0 0 0 − 1 0 0 0 0 0 ] = [ 1 − 1 0 0 0 1 − 1 0 0 0 1 0 0 0 0 0 ] dim ( N ( Σ 2 ) ) = 1 = r 2 → dim V − dim R ( Σ 2 ) = 4 − 3 = 1 \Sigma_2 = \begin{bmatrix} -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_2) \bigg ) = 1 = r_2 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_2)} = 4 - 3 = 1 Σ 2 ​ = ⎣ ⎡ ​ − 1 0 0 0 ​ 1 − 1 0 0 ​ 0 1 − 1 0 ​ 0 0 0 0 ​ ⎦ ⎤ ​ = ⎣ ⎡ ​ 1 0 0 0 ​ − 1 1 0 0 ​ 0 − 1 1 0 ​ 0 0 0 0 ​ ⎦ ⎤ ​ dim ( N ( Σ 2 ​ ) ) = 1 = r 2 ​ → dim V − dim R ( Σ 2 ​ ) = 4 − 3 = 1
Then the minimal polynomial is m ( s ) = ( s − 1 ) 3 ( s − 2 ) m(s) = (s-1)^3(s-2) m ( s ) = ( s − 1 ) 3 ( s − 2 ) .
Example: A = [ 2 1 0 0 0 0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 3 1 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 3 ] A = \begin{bmatrix} 2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix} A = ⎣ ⎡ ​ 2 0 0 0 0 0 0 0 0 0 ​ 1 2 0 0 0 0 0 0 0 0 ​ 0 1 2 0 0 0 0 0 0 0 ​ 0 0 0 2 0 0 0 0 0 0 ​ 0 0 0 1 2 0 0 0 0 0 ​ 0 0 0 0 0 2 0 0 0 0 ​ 0 0 0 0 0 0 2 0 0 0 ​ 0 0 0 0 0 0 0 3 0 0 ​ 0 0 0 0 0 0 0 1 3 0 ​ 0 0 0 0 0 0 0 0 0 3 ​ ⎦ ⎤ ​ d ( s ) = ( s − 2 ) 7 ( s − 3 ) 3 m ( s ) = ? \Large \substack{d(s) = (s-2)^7(s-3)^3 \\ m(s) = ?} d ( s ) = ( s − 2 ) 7 ( s − 3 ) 3 m ( s ) = ? ​
Solution: For each jordan block, we need to check the dimension of N ( Σ 1 i ) N(\Sigma_1^i) N ( Σ 1 i ​ ) .
The largest jordan block will have the largest dimension of N ( Σ 1 i ) N(\Sigma_1^i) N ( Σ 1 i ​ ) . The rest of the jordan blocks will have dimension of N ( Σ 1 i ) N(\Sigma_1^i) N ( Σ 1 i ​ ) equal to the size of the jordan block. Hence the geometric multiplicity wont increase.
Let Σ 1 \Sigma_1 Σ 1 ​ be A − λ 1 I A-\lambda_1I A − λ 1 ​ I and Σ 2 \Sigma_2 Σ 2 ​ be A − λ 2 I A-\lambda_2I A − λ 2 ​ I . Check the largest jordan block.
Σ 1 = [ 0 1 0 0 0 1 0 0 0 ] dim ( N ( Σ 1 ) ) = 1 ≠r 1 → dim V − dim R ( Σ 1 ) = 3 − 2 = 1 \Sigma_1 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1) \bigg ) = 1 \neq r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1)} = 3 - 2 = 1 Σ 1 ​ = ⎣ ⎡ ​ 0 0 0 ​ 1 0 0 ​ 0 1 0 ​ ⎦ ⎤ ​ dim ( N ( Σ 1 ​ ) ) = 1 î€ = r 1 ​ → dim V − dim R ( Σ 1 ​ ) = 3 − 2 = 1
Then we need to check the dimension of N ( Σ 1 2 ) N(\Sigma_1^2) N ( Σ 1 2 ​ ) .
Σ 1 2 = [ 0 0 1 0 0 0 0 0 0 ] dim ( N ( Σ 1 2 ) ) = 2 ≠r 1 → dim V − dim R ( Σ 1 2 ) = 3 − 1 = 2 \Sigma_1^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^2) \bigg ) = 2 \neq r_1 \rightarrow \text{dim}{V}- \text{dim}{R(\Sigma_1^2)} = 3 - 1 = 2 Σ 1 2 ​ = ⎣ ⎡ ​ 0 0 0 ​ 0 0 0 ​ 1 0 0 ​ ⎦ ⎤ ​ dim ( N ( Σ 1 2 ​ ) ) = 2 î€ = r 1 ​ → dim V − dim R ( Σ 1 2 ​ ) = 3 − 1 = 2
Check the dimension of N ( Σ 1 3 ) N(\Sigma_1^3) N ( Σ 1 3 ​ ) .
Σ 1 3 = [ 0 0 0 0 0 0 0 0 0 ] dim ( N ( Σ 1 3 ) ) = 3 = r 1 → dim V − dim R ( Σ 1 3 ) = 3 − 0 = 3 \Sigma_1^3 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^3) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^3)} = 3 - 0 = 3 Σ 1 3 ​ = ⎣ ⎡ ​ 0 0 0 ​ 0 0 0 ​ 0 0 0 ​ ⎦ ⎤ ​ dim ( N ( Σ 1 3 ​ ) ) = 3 = r 1 ​ → dim V − dim R ( Σ 1 3 ​ ) = 3 − 0 = 3
We need to continue to see that further powers of Σ 1 \Sigma_1 Σ 1 ​ will not increase the dimension of N ( Σ 1 i ) N(\Sigma_1^i) N ( Σ 1 i ​ ) .
Σ 1 4 = [ 0 0 0 0 0 0 0 0 0 ] dim ( N ( Σ 1 4 ) ) = 3 = r 1 → dim V − dim R ( Σ 1 4 ) = 3 − 0 = 3 \Sigma_1^4 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^4) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^4)} = 3 - 0 = 3 Σ 1 4 ​ = ⎣ ⎡ ​ 0 0 0 ​ 0 0 0 ​ 0 0 0 ​ ⎦ ⎤ ​ dim ( N ( Σ 1 4 ​ ) ) = 3 = r 1 ​ → dim V − dim R ( Σ 1 4 ​ ) = 3 − 0 = 3
This implies that the further jordan blocks will not increase the dimension of N ( Σ 1 i ) N(\Sigma_1^i) N ( Σ 1 i ​ ) . Hence they are not needed to be checked for minimal polynomial.
Now we can check the second eigenvalue.
Σ 2 = [ 0 − 2 0 0 ] dim ( N ( Σ 2 ) ) = 1 ≠r 2 → dim V − dim R ( Σ 2 ) = 2 − 1 = 1 \Sigma_2 = \begin{bmatrix} 0 & -2 \\ 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_2) \bigg ) = 1 \neq r_2 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_2)} = 2 - 1 = 1 Σ 2 ​ = [ 0 0 ​ − 2 0 ​ ] dim ( N ( Σ 2 ​ ) ) = 1 î€ = r 2 ​ → dim V − dim R ( Σ 2 ​ ) = 2 − 1 = 1
Σ 2 2 = [ 0 0 0 0 ] dim ( N ( Σ 2 2 ) ) = 2 = r 2 → dim V − dim R ( Σ 2 2 ) = 2 − 0 = 2 \Sigma_2^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_2^2) \bigg ) = 2 = r_2 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_2^2)} = 2 - 0 = 2 Σ 2 2 ​ = [ 0 0 ​ 0 0 ​ ] dim ( N ( Σ 2 2 ​ ) ) = 2 = r 2 ​ → dim V − dim R ( Σ 2 2 ​ ) = 2 − 0 = 2
Then the minimal polynomial is m ( s ) = ( s − 2 ) 7 ( s − 3 ) 2 m(s) = (s-2)^7(s-3)^2 m ( s ) = ( s − 2 ) 7 ( s − 3 ) 2 .
Remark:
dim ( N ( A − λ i I ) ) = # of jordan blocks corresponding to λ i  with size ≥  1 \text{dim} \bigg ( N(A-\lambda_iI) \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_i \text{ with size $\geq$ 1}
dim ( N ( A − λ i ​ I ) ) = # of jordan blocks corresponding to λ i ​  with size ≥  1 dim ( N ( A − λ i I ) 2 ) − dim ( N ( A − λ i I ) ) = # of jordan blocks corresponding to λ i  with size ≥  2 \text{dim} \bigg ( N(A-\lambda_iI)^2 \bigg ) - \text{dim} \bigg ( N(A-\lambda_iI) \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_i \text{ with size $\geq$ 2}
dim ( N ( A − λ i ​ I ) 2 ) − dim ( N ( A − λ i ​ I ) ) = # of jordan blocks corresponding to λ i ​  with size ≥  2 dim ( N ( A − λ i I ) 3 ) − dim ( N ( A − λ i I ) 2 ) = # of jordan blocks corresponding to λ i  with size ≥  3 \text{dim} \bigg ( N(A-\lambda_iI)^3 \bigg ) - \text{dim} \bigg ( N(A-\lambda_iI)^2 \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_i \text{ with size $\geq$ 3}
dim ( N ( A − λ i ​ I ) 3 ) − dim ( N ( A − λ i ​ I ) 2 ) = # of jordan blocks corresponding to λ i ​  with size ≥  3 dim ( N ( A − λ i I ) k ) − dim ( N ( A − λ i I ) k − 1 ) = # of jordan blocks corresponding to λ i  with size ≥  k \text{dim} \bigg ( N(A-\lambda_iI)^k \bigg ) - \text{dim} \bigg ( N(A-\lambda_iI)^{k-1} \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_i \text{ with size $\geq$ k}
dim ( N ( A − λ i ​ I ) k ) − dim ( N ( A − λ i ​ I ) k − 1 ) = # of jordan blocks corresponding to λ i ​  with size ≥  k
Example: A ∈ R 4 A \in \mathbb{R}^4 A ∈ R 4 Which has a single eigenvalue λ 1 = 7 \lambda_1 = 7 λ 1 ​ = 7 and the geometric multiplicity is 2. Find all possible jordan forms.
Solution: The characteristic polynomial is d ( s ) = ( s − 7 ) 4 d(s) = (s-7)^4 d ( s ) = ( s − 7 ) 4 and the minimal polynomial is m ( s ) = ( s − 7 ) 2 m(s) = (s-7)^2 m ( s ) = ( s − 7 ) 2 . This suggests that,
dim ( N ( A − 7 I ) 2 ) = 4 = r 1 → dim ( V ) = dim ( N ( A − 7 I ) 2 ) \text{dim} \bigg ( N(A-7I)^2 \bigg ) = 4 = r_1 \rightarrow \text{dim}(V) = \text{dim} \bigg ( N(A-7I)^2 \bigg ) dim ( N ( A − 7 I ) 2 ) = 4 = r 1 ​ → dim ( V ) = dim ( N ( A − 7 I ) 2 )
Then for the dim ( N ( A − 7 I ) ) \text{dim}\bigg (N(A-7I)\bigg ) dim ( N ( A − 7 I ) ) we have 3 possibilities, 1, 2, or 3.
a) Let dim N ( A − 7 I ) = 3 \text{a) Let } \text{dim} N(A-7I) = 3 a) Let dim N ( A − 7 I ) = 3
     # of jordan blocks = 3 \ \ \ \ \ \text{\# of jordan blocks } = 3      # of jordan blocks = 3
     # of jordan blocks w/ size ≥ 2 = 4 − 3 = 1 \ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 2 = 4 - 3 = 1      # of jordan blocks w/ size ≥ 2 = 4 − 3 = 1
     # of jordan blocks w/ size ≥ 3 = 3 − 3 = 0 \ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 3 = 3 - 3 = 0      # of jordan blocks w/ size ≥ 3 = 3 − 3 = 0
A = [ 7 1 0 0 0 7 0 0 0 0 7 0 0 0 0 7 ] A = \begin{bmatrix} 7 & 1 & 0 & 0 \\ 0 & 7 & 0 & 0 \\ 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & 7 \end{bmatrix} A = ⎣ ⎡ ​ 7 0 0 0 ​ 1 7 0 0 ​ 0 0 7 0 ​ 0 0 0 7 ​ ⎦ ⎤ ​
a) Let dim N ( A − 7 I ) = 2 \text{a) Let } \text{dim} N(A-7I) = 2 a) Let dim N ( A − 7 I ) = 2
     # of jordan blocks = 2 \ \ \ \ \ \text{\# of jordan blocks } = 2      # of jordan blocks = 2
     # of jordan blocks w/ size ≥ 2 = 4 − 2 = 2 \ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 2 = 4 - 2 = 2      # of jordan blocks w/ size ≥ 2 = 4 − 2 = 2
     # of jordan blocks w/ size ≥ 3 = dim ( N ( A − 7 I ) 3 ) −  dim ( N ( A − 7 I ) 2 ) = 4 − 4 = 0 \ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 3 = \text{dim}\bigg ( N(A-7I)^3 \bigg ) - \text{ dim}\bigg ( N(A-7I)^2 \bigg ) = 4 - 4 = 0      # of jordan blocks w/ size ≥ 3 = dim ( N ( A − 7 I ) 3 ) −  dim ( N ( A − 7 I ) 2 ) = 4 − 4 = 0
A = [ 7 1 0 0 0 7 0 0 0 0 7 1 0 0 0 7 ] A = \begin{bmatrix} 7 & 1 & 0 & 0 \\ 0 & 7 & 0 & 0 \\ 0 & 0 & 7 & 1 \\ 0 & 0 & 0 & 7 \end{bmatrix} A = ⎣ ⎡ ​ 7 0 0 0 ​ 1 7 0 0 ​ 0 0 7 0 ​ 0 0 1 7 ​ ⎦ ⎤ ​
a) Let dim N ( A − 7 I ) = 1 \text{a) Let } \text{dim} N(A-7I) = 1 a) Let dim N ( A − 7 I ) = 1
     # of jordan blocks = 1 \ \ \ \ \ \text{\# of jordan blocks } = 1      # of jordan blocks = 1
     # of jordan blocks w/ size ≥ 2 = 4 − 1 = 3 \ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 2 = 4 - 1 = 3      # of jordan blocks w/ size ≥ 2 = 4 − 1 = 3
However, this is not possible since there cannot be 3 jordan blocks with size greater than or equal to 2.
Example: Suppose a matrix A ∈ R 8 × 8 A \in \mathbb{R}^{8\times 8} A ∈ R 8 × 8 has the following subspace dimensions:
dim ( N ( A − 3 I ) ) = 5 dim ( N ( A − 3 I ) 2 ) = 7 dim ( N ( A − 3 I ) 3 ) = 8 \text{dim} \bigg ( N(A-3I) \bigg ) = 5 \\
\text{dim} \bigg ( N(A-3I)^2 \bigg ) = 7 \\
\text{dim} \bigg ( N(A-3I)^3 \bigg ) = 8 dim ( N ( A − 3 I ) ) = 5 dim ( N ( A − 3 I ) 2 ) = 7 dim ( N ( A − 3 I ) 3 ) = 8
a) Find the characteristic polynomial of A.
b) Find the minimal polynomial of A.
c) Find the possible Jordan forms of A.
Solution:
a) We know that the characteristic polynomial is d ( s ) = ( s − 3 ) 8 d(s) = (s-3)^8 d ( s ) = ( s − 3 ) 8 .
b) We know that the minimal polynomial is m ( s ) = ( s − 3 ) 3 m(s) = (s-3)^3 m ( s ) = ( s − 3 ) 3 since the dimension of N ( A − 3 I ) 3 N(A-3I)^3 N ( A − 3 I ) 3 is same as the dimension of the vector space. This implies that the geometric multiplicity is 3.
c) Starting from the largest jordan block, having geometric multiplicity 3 suggest that the largest jordan block has size 3.
dim ( N ( A − 3 I ) 3 ) − dim ( N ( A − 3 I ) 2 ) = # of jordan blocks corresponding to λ 1  with size ≥ 3 = 8 − 7 = 1 \text{dim} \bigg ( N(A-3I)^3 \bigg ) - \text{dim} \bigg ( N(A-3I)^2 \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_1 \text{ with size }\geq 3 = 8 - 7 = 1 dim ( N ( A − 3 I ) 3 ) − dim ( N ( A − 3 I ) 2 ) = # of jordan blocks corresponding to λ 1 ​  with size ≥ 3 = 8 − 7 = 1
dim ( N ( A − 3 I ) 2 ) − dim ( N ( A − 3 I ) ) = # of jordan blocks corresponding to λ 1  with size ≥ 2 = 7 − 5 = 2 \text{dim} \bigg ( N(A-3I)^2 \bigg ) - \text{dim} \bigg ( N(A-3I) \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_1 \text{ with size }\geq 2 = 7 - 5 = 2 dim ( N ( A − 3 I ) 2 ) − dim ( N ( A − 3 I ) ) = # of jordan blocks corresponding to λ 1 ​  with size ≥ 2 = 7 − 5 = 2
We have 2 that are greater than or equal to 2. This implies that we have 1 jordan block with size 3 and 1 jordan blocks with size 2.
dim ( N ( A − 3 I ) ) = # of jordan blocks corresponding to λ 1  with size ≥ 1 = 5 \text{dim} \bigg ( N(A-3I) \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_1 \text{ with size }\geq 1 = 5 dim ( N ( A − 3 I ) ) = # of jordan blocks corresponding to λ 1 ​  with size ≥ 1 = 5
This implies that we have 3 jordan blocks with size 1.
A = [ 3 1 0 0 0 0 0 0 0 3 1 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 3 1 0 0 0 0 0 0 0 3 1 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 3 1 0 0 0 0 0 0 0 3 ] A = \begin{bmatrix}3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix} A = ⎣ ⎡ ​ 3 0 0 0 0 0 0 0 ​ 1 3 0 0 0 0 0 0 ​ 0 1 3 0 0 0 0 0 ​ 0 0 0 3 0 0 0 0 ​ 0 0 0 1 3 0 0 0 ​ 0 0 0 0 1 3 0 0 ​ 0 0 0 0 0 0 3 0 ​ 0 0 0 0 0 0 1 3 ​ ⎦ ⎤ ​
#EE501 - Linear Systems Theory at METU